Light Pressure Modified Orbits

Light pressure effects modify server-sat array orbits. In the nominal orbit, server-sats are at "half thrust", with each thruster half mirror and half transparent. Variations to full or zero reflectivity, and full or zero thrust, allow each server-sat to maneuver in relation to the array, or for the array as a whole to maneuver around its assigned centerpoint. The following is an analysis of two effects, earth oblateness and nominal half-thrust light pressure, on the orbit. We will assume that the arrays maintain a constant, slightly elliptical orbit that precesses once per year in the equatorial plane. We will assume continuous illumination tangential to the equatorial plane, and zero light pressure effects from Earth albedo or black-body radiation, and no solar or lunar tides. These assumptions are somewhat crude approximations to get us into the ballpark of a solution. Precise solutions will probably demand accurate numerical solutions simulating many years of orbital evolution.

Earth Oblateness

For "heavy" server-sats relatively close to the earth, such as 3 gram satellites at m288, the dominant deviation from a perfect Kepler orbit is caused by the J_2 spherical harmonic of the gravity field, in turn caused by the oblateness of the spinning Earth. For small eccentricities, the eastward precession of the perigee of one elliptical equatorial orbit is proportional to the J_2 term of the WGS84 model ( -1.082626683E-03 , see Pisacane 2008 ) and the inverse of the orbit radius squared. The precession, expressed as the number complete precessions per year, is J \approx -3 J_2 ( Y / P ) ( R_e / R_s ) ^ 2 ~ ~ where

Light Pressure

Previously I was not using the correct math for fictitious forces in a rotating frame. http://en.wikipedia.org/wiki/Rotating_frame has a good discussion, although they use \Omega where I use \omega_{m288} .

I will show that for 3 gram 50 micron thick flat-sats in m288 orbits, the orbital perturbations from light pressure are small, 180 meters front to back oscillations along the line of the orbit at the spring and fall equinoxes, and 165 meters front to back at the summer and winter solstices. If the satellites get thinner, the perturbations increase proportional to the area to mass ratio. If the orbits move further out, the perturbations increase proportional to the cube of the orbit radius, because the orbit period grows, and the perturbations are proportional to the square of the orbit period. There is also a cumulative perturbation caused by the eclipse time, which is probably small enough to be corrected by optical maneuvering.

The fictitious forces are proportional to the radial and tangential position from mean perturbation center, and the tangential velocity relative to that center. Assume a circular orbit - toroidal orbits do not deviate much from that.

An m288 orbit has the following parameters (subject to verification, please help me check them):

In a circular orbit, the centrifugal acceleration balances the centripedal gravitational acceleration:

v^2 / r = \omega^2 r = a = \mu_{\oplus} / r^2 ~ ~ ~ ~ = 2.4371020 m/s^2 at m288

\omega^2 = \mu_{\oplus} / r^3

If x \equiv the tangential distance forward of orbit center, then for small x the triangle of tangential and radial accelerations is proportional to the tangential and radial distances, from congruent triangles:

\partial a_x / a = - \partial x / r

a_x = -( a / r ) x = - \omega^2 x

With no perturbations, the vertical acceleration is:

a_r = \omega^2 ~ r - \mu_{\oplus} / r^2 = v^2 / r - \mu_{\oplus} / r^2 = 0

If the tangential velocity is perturbed, the radial acceleration is perturbed:

\partial a_r = 2 v / r \partial v_x = 2 \omega ~ \partial v_x

If the radial distance is perturbed, the radial acceleration is also perturbed:

\partial a_r = ( \omega^2 + 2 \mu_{\oplus} / r^3 ) \partial r = 3 \omega^2 \partial r

So the total radial acceleration, for small perturbations of y \equiv \Delta r and x is:

a_y = 3 \omega^2 ~ y + 2 \omega ~ v_x

The radial and tangential accelerations are caused by the light pressure from the Sun. This can be divided into two components, the planar light pressure parallel to the plane and the light pressure perpendicular to the plane. These are a function of the time of year \beta , and the Earth's axial tilt of \phi = 23.439281° . We can compute the components from the cross product of the sunwards light pressure vector a_{\lambda} ~ \hat j and the normal vector of the equatorial plane given by \hat i = \sin( \phi ) \cos( \beta ) , \hat j = \sin( \phi ) \sin( \beta ) , and \hat k = \cos ( \phi ) . The perpendicular pressure is a_{\lambda ~ z} = a_{\lambda} \sin( \phi ) \sin( \beta ) , and the planar pressure is a_{\lambda ~ p} = a_{\lambda} sqrt{ 1 - ( \sin( \phi ) \sin( \beta ) )^2 } . At the equinoxes, a_{\lambda ~ z} = 0 and a_{\lambda ~ p} = a_{\lambda} . At the solstices, a_{\lambda ~ z} = \pm a_{\lambda} \sin( \phi ) and a_{\lambda ~ p} = a_{\lambda} \cos( \phi ) $.

The perpendicular component does not change as the object orbits. The object is displaced above or below the equatorial plane by z ~ = ~ a_{\lambda ~ z} / \omega^{2 ~ = ~ ( a_{\lambda} / { \omega}2 } ) \sin ( \phi ) \sin( \beta ) . Define the parameter \lambda ~ \equiv ~ a_{\lambda} / { \omega^2 } = 180.3 meters for a 3 gram, 50 \mum thick glass flat sat at m288 . The z displacement varies sinusoidally between \pm$ 71.7 meters over the course of a year, far smaller than the cross section of the toroidal orbit.

\omega^2 = \mu_{\oplus} / r^3 , so \lambda = a_{\lambda} r^3 / \mu_{\oplus} . A thinner server satellite has higher a_{\lambda} and higher \lambda . A more distant orbit also has higher \lambda . For sufficiently thin satellites or large distances, the approximations above break down, and light pressure will push the satellites out of orbit, perhaps to escape velocity.

The planar light pressure makes one rotation per 14400 seconds around the guiding center, and is interrupted when the satellite is eclipsed by the Earth. We will compute the effect of this later. If we approximate the rotation as the sidereal period instead, and assume we can make up for eclipse and rotation changes by maneuvering (dangerous assumption), then we can approximate the light pressure components as:

a_{{\lambda} ~ y } = a_{\lambda ~ p} \sin( \omega t )

a_{{\lambda} ~ x } = - a_{\lambda ~ p } \cos( \omega t )

Assume that x = k \lambda \cos( \omega t ) ~ ~

Then ~ ~ ~ v_x = \dot { x } = - k \lambda \omega \sin( \omega t )

and ~ ~ ~ \ddot { x } = - k \lambda \omega^2 \cos( \omega t ) = - \omega^2 x

The sum of the inertial force and the displacement force is equal to the light pressure:

- a_{\lambda ~ p } \cos( \omega t ) = \ddot { x } + a_x = - \omega^2 x + - \omega^2 x = -2 \omega^2 k \lambda \cos( \omega t )

Dividing both sides by \omega^2 \cos( \omega t ) :

- a_{\lambda ~ p } / \omega^2 = \lambda = -2 k \lambda . Thus, k = -0.5 .

Therefore x = - 0.5 \lambda \cos( \omega t ) and v_x = - 0.5 a_{\lambda} / \omega \sin( \omega t )

The acceleration of y is the a_y fictitious force plus the light pressure:

\ddot{ y } = a_y + a_{{\lambda ~ p } ~ y } = 3 \omega^2 y + 2 \omega v_x + a_{\lambda ~ p} \sin( \omega t ) = 3 \omega^2 y + 2 \omega ( - a_{\lambda ~ p} / ( 2 \omega ) ) \sin( \omega t ) + a_{\lambda ~ p} \sin( \omega t ) = 3 \omega^2 y - a_{\lambda ~ p} \sin( \omega t ) + a_{\lambda ~ p} \sin( \omega t ) = 3 \omega^2 y

If \ddot{ y } = 3 \omega^2 y, the only stable solution is y = 0 !!! The satellite oscillates back and forth along the path of the orbit, and is displaced perpendicular to the orbit, but is not displaced radially by light pressure. The centrifugal acceleration of the tangential oscillation exactly balances the light pressure.

Perturbations because of eclipse

The above would be accurate if the earth was transparent. However, the satellite passes behind the earth for as much as 16.1% of its orbit during the equinoxes, and 11.1% at the solstices.

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